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B (Salters) F331 (Chemistry for Life) 22nd May 2015 [Official Thread]

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Original post by vuvuzela
Hess Law will blow up this year too.
So will the three hydrocarbon processes

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what do you mean by that?)
can someone explain the calculations of % on page 42 of the test bank pls
Original post by TheLegalDealer
Divide the percentage of each atom which is basically its mass in the compound by its atmoic mass to find the moles of it

Do this for all the atoms in the compound .Now find the ratio of the atoms by dividing each of the atoms moles by the moles of the atom that he smallest mole

This would give you a ratio e.g. 2:2:1. Then just put the ratio of each atom in the formula e.g. C2H2O


thanku :smile:
Can anyone explain Question 2d on the june 2014 paper its doing my head in
When given bond enthalpies, to work out enthalpy change, do you do products - reactants?
or reactants - products?
Reply 65
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barbiex
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Original post by TheLegalDealer
Can anyone explain Question 2d on the june 2014 paper its doing my head in


same here, still cant get the correct answer after like 10 attempts wtf, helppppppp
Sulfur is 10% of 1kg, so mass of sulfur is 100g, and Mr is 32.1
Mol=mass/mr, so you can find moles of sulfur
I assumed moles of sulfur is in the same ratio as mole of SO2 so calculated mole is mole of SO2
Then you can rearrange formula so that mass= moles x Mr of S02, which gives you about 200g
Original post by AAlice97
Sulfur is 10% of 1kg, so mass of sulfur is 100g, and Mr is 32.1
Mol=mass/mr, so you can find moles of sulfur
I assumed moles of sulfur is in the same ratio as mole of SO2 so calculated mole is mole of SO2
Then you can rearrange formula so that mass= moles x Mr of S02, which gives you about 200g


thank you so much !!!
Reply 68
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dansag
1
can someone please explain hess law
Original post by dansag
can someone please explain hess law


If the starting and finishing points are the same the enthalpy change for a chemical reaction will always be the same regardless of the route it takes
Isn't it starting and finishing conditions must be the same not points ? how specific is the wording ?
Reply 71
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vuvuzela
OP
3
Original post by Obereicher
Isn't it starting and finishing conditions must be the same not points ? how specific is the wording ?


Not that specific, the previous reply is accurate and would get you all marks

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Original post by buckeybarnes
can someone explain the calculations of % on page 42 of the test bank pls


1.75% of 100g = 1.75g

n=m/Mr

1.75g/9= 0.19 moles (Be)

100g-1.75g= 98.25g

n= m/Mr

98.25/63.25 = 1.55 moles (Cu)

:smile:
(edited 8 years ago)
Original post by vuvuzela
Not that specific, the previous reply is accurate and would get you all marks

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Screen Shot 05-21-15 at 10.02 PM 001.PNG

It would seem you do unfortunately :frown:
Reply 74
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Bruhh
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Original post by Obereicher
Isn't it starting and finishing conditions must be the same not points ? how specific is the wording ?


Original post by vuvuzela
Not that specific, the previous reply is accurate and would get you all marks

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No you have to have conditions, it says not to accept points even though that's what the revision guide says which is strange.
Reply 75
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vuvuzela
OP
3
^ i stand corrected. At least i know for tomorrow and hope it comes up

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Original post by chococup123
When given bond enthalpies, to work out enthalpy change, do you do products - reactants?
or reactants - products?


You do reactants- products

If its negative= exothermic
positive= endothermic
Original post by vuvuzela
Hess Law will blow up this year too.
So will the three hydrocarbon processes

Posted from TSR Mobile


What do u mean by the 3 hydrocarbon processes?

Is it isomerism, cracking and reforming?
Reply 78
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vuvuzela
OP
3
Original post by shooting_stars
What do u mean by the 3 hydrocarbon processes?

Is it isomerism, cracking and reforming?


Yeah. Good luck all

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Last min prep guys , what most likely to come up ??

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